How do you solve this challenging function below?

Question

Let f(x) = 2x^2 - 7x + 3 and g(x) = 2x^2 - 9x + 9.

Then

lim..... f(x)/g(x) =
x=>-3


The book gives these possible answers...

a. 5/3
b. 4/3
c. 1
d. 2/3
e. 7/9
f. 4/9
g. 5/7
h. 4/7
i. 3/4
j. 9/4
k. 9/10
l. none of these

Answer 1

e) plug x = -3 into the equation

Answer 2

I think there is a misprint in your book, or I have misunderstood your question! (I'm reading it as "the limit of f(x)/g(x) as x approaches -3 from above.") Because if you substitute x=-3 into f(x)/g(x) you do get an answer (ie f(-3)/g(-3) is defined), therefore the limit part of the question is redundant.


You can simplify f(x)/g(x) as follows~

First of all, f(x)/g(x) is a "top-heavy" fraction because the degree of the numerator is the same as the degree of the denominator. So you can split it into a whole number plus a fraction:

f(x)/g(x) = 1 + (2x-6)/(2x^2-9x+9)

Now factorise the top and bottom of the fraction:

= 1 + 2(x-3)/(2x-3)(x-3)

Cancel common factors:

= 1 + 2/(2x-3)


As this function is therefore undefined for x = 1.5 I think it more likely you are supposed to find the limit as x approaches 1.5 from above. In which case:

as x approaches 1.5 from above,
(2x-3) approaches zero from above,
so 2/(2x-3) approaches + infinity
so 1+2/(2x-3) approaches + infinity as well.

However if a approaches 1.5 from below, the whole thing approaches - infinity.

Answer 3

e) 7/9

Answer 4

lim..... f(x)/g(x) =
x=>-3

lim (2x² - 7x + 3) / (2x² - 9x + 9) =
x=>-3

lim d [(2x² - 7x + 3) / (2x² - 9x + 9)] /dx =
x=>-3

lim [(4x - 7) / (4x - 9)] /dx = (4*3 - 7) / (4*3 - 9)]
x=>-3

= (4*3 - 7) / (4*3 - 9) = 5/3

answer: letter (a)

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